Multiple models in a view

Question

I want to have 2 models in one view. The page contains both LoginViewModel and RegisterViewModel.

e.g.

public class LoginViewModel
{
    public string Email { get; set; }
    public string Password { get; set; }
}

public class RegisterViewModel
{
    public string Name { get; set; }
    public string Email { get; set; }
    public string Password { get; set; }
}

Do I need to make another ViewModel which holds these 2 ViewModels?

public BigViewModel
{
    public LoginViewModel LoginViewModel{get; set;}
    public RegisterViewModel RegisterViewModel {get; set;}
}

I need the validation attributes to be brought forward to the view. This is why I need the ViewModels.

Isn't there another way such as (without the BigViewModel):

 @model ViewModel.RegisterViewModel
 @using (Html.BeginForm("Login", "Auth", FormMethod.Post))
 {
        @Html.TextBoxFor(model => model.Name)
        @Html.TextBoxFor(model => model.Email)
        @Html.PasswordFor(model => model.Password)
 }

 @model ViewModel.LoginViewModel
 @using (Html.BeginForm("Login", "Auth", FormMethod.Post))
 {
        @Html.TextBoxFor(model => model.Email)
        @Html.PasswordFor(model => model.Password)
 }

Answer 1

There are lots of ways...

1. with your BigViewModel you do:

   @model BigViewModel    
   @using(Html.BeginForm()) {
       @Html.EditorFor(o => o.LoginViewModel.Email)
       ...
   }
   

2. you can create 2 additional views

   Login.cshtml

   @model ViewModel.LoginViewModel
   @using (Html.BeginForm("Login", "Auth", FormMethod.Post))
   {
       @Html.TextBoxFor(model => model.Email)
       @Html.PasswordFor(model => model.Password)
   }
   

   and register.cshtml same thing

   after creation you have to render them in the main view and pass them the viewmodel/viewdata

   so it could be like this:

   @{Html.RenderPartial("login", ViewBag.Login);}
   @{Html.RenderPartial("register", ViewBag.Register);}
   

   or

   @{Html.RenderPartial("login", Model.LoginViewModel)}
   @{Html.RenderPartial("register", Model.RegisterViewModel)}
   

3. using ajax parts of your web-site become more independent

4. iframes, but probably this is not the case

Answer 2

I'd recommend using Html.RenderAction and PartialViewResults to accomplish this; it will allow you to display the same data, but each partial view would still have a single view model and removes the need for a BigViewModel

So your view contain something like the following:

@Html.RenderAction("Login")
@Html.RenderAction("Register")

Where Login & Register are both actions in your controller defined like the following:

public PartialViewResult Login( )
{
    return PartialView( "Login", new LoginViewModel() );
}

public PartialViewResult Register( )
{
    return PartialView( "Register", new RegisterViewModel() );
}

The Login & Register would then be user controls residing in either the current View folder, or in the Shared folder and would like something like this:

/Views/Shared/Login.cshtml: (or /Views/MyView/Login.cshtml)

@model LoginViewModel
@using (Html.BeginForm("Login", "Auth", FormMethod.Post))
{
    @Html.TextBoxFor(model => model.Email)
    @Html.PasswordFor(model => model.Password)
}

/Views/Shared/Register.cshtml: (or /Views/MyView/Register.cshtml)

@model ViewModel.RegisterViewModel
@using (Html.BeginForm("Login", "Auth", FormMethod.Post))
{
    @Html.TextBoxFor(model => model.Name)
    @Html.TextBoxFor(model => model.Email)
    @Html.PasswordFor(model => model.Password)
}

And there you have a single controller action, view and view file for each action with each totally distinct and not reliant upon one another for anything.

Answer 3

Another way is to use:

@model Tuple<LoginViewModel,RegisterViewModel>

I have explained how to use this method both in the view and controller for another example: Two models in one view in ASP MVC 3

In your case you could implement it using the following code:

In the view:

@using YourProjectNamespace.Models;
@model Tuple<LoginViewModel,RegisterViewModel>

@using (Html.BeginForm("Login1", "Auth", FormMethod.Post))
{
        @Html.TextBoxFor(tuple => tuple.Item2.Name, new {@Name="Name"})
        @Html.TextBoxFor(tuple => tuple.Item2.Email, new {@Name="Email"})
        @Html.PasswordFor(tuple => tuple.Item2.Password, new {@Name="Password"})
}

@using (Html.BeginForm("Login2", "Auth", FormMethod.Post))
{
        @Html.TextBoxFor(tuple => tuple.Item1.Email, new {@Name="Email"})
        @Html.PasswordFor(tuple => tuple.Item1.Password, new {@Name="Password"})
}

Note that I have manually changed the Name attributes for each property when building the form. This needs to be done, otherwise it wouldn't get properly mapped to the method's parameter of type model when values are sent to the associated method for processing. I would suggest using separate methods to process these forms separately, for this example I used Login1 and Login2 methods. Login1 method requires to have a parameter of type RegisterViewModel and Login2 requires a parameter of type LoginViewModel.

if an actionlink is required you can use:

@Html.ActionLink("Edit", "Edit", new { id=Model.Item1.Id })

in the controller's method for the view, a variable of type Tuple needs to be created and then passed to the view.

Example:

public ActionResult Details()
{
    var tuple = new Tuple<LoginViewModel, RegisterViewModel>(new LoginViewModel(),new RegisterViewModel());
    return View(tuple);
}

or you can fill the two instances of LoginViewModel and RegisterViewModel with values and then pass it to the view.

Answer 4

Use a view model that contains multiple view models:

   namespace MyProject.Web.ViewModels
   {
      public class UserViewModel
      {
          public UserDto User { get; set; }
          public ProductDto Product { get; set; }
          public AddressDto Address { get; set; }
      }
   }

In your view:

  @model MyProject.Web.ViewModels.UserViewModel

  @Html.LabelFor(model => model.User.UserName)
  @Html.LabelFor(model => model.Product.ProductName)
  @Html.LabelFor(model => model.Address.StreetName)

Answer 5

Do I need to make another view which holds these 2 views?

Answer:No

Isn't there another way such as (without the BigViewModel):

Yes, you can use Tuple (brings magic in view having multiple model).

Code:

 @model Tuple<LoginViewModel, RegisterViewModel>


    @using (Html.BeginForm("Login", "Auth", FormMethod.Post))
    {
     @Html.TextBoxFor(tuple=> tuple.Item.Name)
     @Html.TextBoxFor(tuple=> tuple.Item.Email)
     @Html.PasswordFor(tuple=> tuple.Item.Password)
    }


    @using (Html.BeginForm("Login", "Auth", FormMethod.Post))
     {
      @Html.TextBoxFor(tuple=> tuple.Item1.Email)
      @Html.PasswordFor(tuple=> tuple.Item1.Password)
     }

Answer 6

Add this ModelCollection.cs to your Models

using System;
using System.Collections.Generic;

namespace ModelContainer
{
  public class ModelCollection
  {
   private Dictionary<Type, object> models = new Dictionary<Type, object>();

   public void AddModel<T>(T t)
   {
      models.Add(t.GetType(), t);
   }

   public T GetModel<T>()
   {
     return (T)models[typeof(T)];
   }
 }
}

Controller:

public class SampleController : Controller
{
  public ActionResult Index()
  {
    var model1 = new Model1();
    var model2 = new Model2();
    var model3 = new Model3();

    // Do something

    var modelCollection = new ModelCollection();
    modelCollection.AddModel(model1);
    modelCollection.AddModel(model2);
    modelCollection.AddModel(model3);
    return View(modelCollection);
  }
}

The View:

enter code here
@using Models
@model ModelCollection

@{
  ViewBag.Title = "Model1: " + ((Model.GetModel<Model1>()).Name);
}

<h2>Model2: @((Model.GetModel<Model2>()).Number</h2>

@((Model.GetModel<Model3>()).SomeProperty

Answer 7

a simple way to do that

we can call all model first

@using project.Models

then send your model with viewbag

// for list
ViewBag.Name = db.YourModel.ToList();

// for one
ViewBag.Name = db.YourModel.Find(id);

and in view

// for list
List<YourModel> Name = (List<YourModel>)ViewBag.Name ;

//for one
YourModel Name = (YourModel)ViewBag.Name ;

then easily use this like Model

Answer 8

My advice is to make a big view model:

public BigViewModel
{
    public LoginViewModel LoginViewModel{get; set;}
    public RegisterViewModel RegisterViewModel {get; set;}
}

In your Index.cshtml, if for example you have 2 partials:

@addTagHelper *,Microsoft.AspNetCore.Mvc.TagHelpers
@model .BigViewModel

@await Html.PartialAsync("_LoginViewPartial", Model.LoginViewModel)

@await Html.PartialAsync("_RegisterViewPartial ", Model.RegisterViewModel )

and in controller:

model=new BigViewModel();
model.LoginViewModel=new LoginViewModel();
model.RegisterViewModel=new RegisterViewModel(); 

Answer 9

I want to say that my solution was like the answer provided on this stackoverflow page: ASP.NET MVC 4, multiple models in one view?

However, in my case, the linq query they used in their Controller did not work for me.

This is said query:

var viewModels = 
        (from e in db.Engineers
         select new MyViewModel
         {
             Engineer = e,
             Elements = e.Elements,
         })
        .ToList();

Consequently, "in your view just specify that you're using a collection of view models" did not work for me either.

However, a slight variation on that solution did work for me. Here is my solution in case this helps anyone.

Here is my view model in which I know I will have just one team but that team may have multiple boards (and I have a ViewModels folder within my Models folder btw, hence the namespace):

namespace TaskBoard.Models.ViewModels
{
    public class TeamBoards
    {
        public Team Team { get; set; }
        public List<Board> Boards { get; set; }
    }
}

Now this is my controller. This is the most significant difference from the solution in the link referenced above. I build out the ViewModel to send to the view differently.

public ActionResult Details(int? id)
        {
            if (id == null)
            {
                return new HttpStatusCodeResult(HttpStatusCode.BadRequest);
            }

            TeamBoards teamBoards = new TeamBoards();
            teamBoards.Boards = (from b in db.Boards
                                 where b.TeamId == id
                                 select b).ToList();
            teamBoards.Team = (from t in db.Teams
                               where t.TeamId == id
                               select t).FirstOrDefault();

            if (teamBoards == null)
            {
                return HttpNotFound();
            }
            return View(teamBoards);
        }

Then in my view I do not specify it as a list. I just do "@model TaskBoard.Models.ViewModels.TeamBoards" Then I only need a for each when I iterate over the Team's boards. Here is my view:

@model TaskBoard.Models.ViewModels.TeamBoards

@{
    ViewBag.Title = "Details";
}

<h2>Details</h2>

<div>
    <h4>Team</h4>
    <hr />


    @Html.ActionLink("Create New Board", "Create", "Board", new { TeamId = @Model.Team.TeamId}, null)
    <dl class="dl-horizontal">
        <dt>
            @Html.DisplayNameFor(model => Model.Team.Name)
        </dt>

        <dd>
            @Html.DisplayFor(model => Model.Team.Name)
            <ul>
                @foreach(var board in Model.Boards)
                { 
                    <li>@Html.DisplayFor(model => board.BoardName)</li>
                }
            </ul>
        </dd>

    </dl>
</div>
<p>
    @Html.ActionLink("Edit", "Edit", new { id = Model.Team.TeamId }) |
    @Html.ActionLink("Back to List", "Index")
</p>

I am fairly new to ASP.NET MVC so it took me a little while to figure this out. So, I hope this post helps someone figure it out for their project in a shorter timeframe. :-)

Answer 10

1. Create one new class in your model and properties of LoginViewModel and RegisterViewModel:

   public class UserDefinedModel() 
   {
       property a1 as LoginViewModel 
       property a2 as RegisterViewModel 
   }
   

2. Then use UserDefinedModel in your view.

Answer 11

you can always pass the second object in a ViewBag or View Data.

Answer 12

This is a simplified example with IEnumerable.

I was using two models on the view: a form with search criteria (SearchParams model), and a grid for results, and I struggled with how to add the IEnumerable model and the other model on the same view. Here is what I came up with, hope this helps someone:

@using DelegatePortal.ViewModels;

@model SearchViewModel

@using (Html.BeginForm("Search", "Delegate", FormMethod.Post))
{

                Employee First Name
                @Html.EditorFor(model => model.SearchParams.FirstName,
new { htmlAttributes = new { @class = "form-control form-control-sm " } })

                <input type="submit" id="getResults" value="SEARCH" class="btn btn-primary btn-lg btn-block" />

}
<br />
    @(Html
        .Grid(Model.Delegates)
        .Build(columns =>
        {
            columns.Add(model => model.Id).Titled("Id").Css("collapse");
            columns.Add(model => model.LastName).Titled("Last Name");
            columns.Add(model => model.FirstName).Titled("First Name");
        })

... )

SearchViewModel.cs:

namespace DelegatePortal.ViewModels
{
    public class SearchViewModel
    {
        public IEnumerable<DelegatePortal.Models.DelegateView> Delegates { get; set; }

        public SearchParamsViewModel SearchParams { get; set; }
....

DelegateController.cs:

// GET: /Delegate/Search
    public ActionResult Search(String firstName)
    {
        SearchViewModel model = new SearchViewModel();
        model.Delegates = db.Set<DelegateView>();
        return View(model);
    }

    // POST: /Delegate/Search
    [HttpPost]
    public ActionResult Search(SearchParamsViewModel searchParams)
    {
        String firstName = searchParams.FirstName;
        SearchViewModel model = new SearchViewModel();

        if (firstName != null)
            model.Delegates = db.Set<DelegateView>().Where(x => x.FirstName == firstName);

        return View(model);
    }

SearchParamsViewModel.cs:

namespace DelegatePortal.ViewModels
{
    public class SearchParamsViewModel
    {
        public string FirstName { get; set; }
    }
}